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    Given that m is an odd positive interger, prove that (m2 + 3)(m2 + 15) is divisible by 32 for all such values of m.
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    m=2k+1, k>=0

    (m^2 + 3)(m^2 + 15)
    <=>
    ((2k+1)^2 + 3)((2k+1)^2 + 15)
    <=>
    (4k^2+4k+4)(4k^2+4k+16)=16(k^2+k +1)(k^2+k+4)

    Either (k^2+k+1) or (k^2+k+4) will be even, and thus divisible by 2.

    2*16 = 32. And we're done.

    It really isn't necessary to do it by induction. But it can be made in a similar fashion by considering the case k=1 and the case k=p+1, and in the latter use the same reasoning as above.
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    Uh, actually, k^2+k+4 will always be even.

    E.g. consider k=2m+a, m>=0 and a is either 0 or 1, yields
    k^2+k+4 = (2m+a)^2+2m+a = 2(2m^2+2ma+m) + a^2+a.
    a^2+a = 2 or 0.
 
 
 
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Updated: April 9, 2006
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